The function f(x)=4 ^3x-6 ^2x+12 +100 is strictly: - Vidyadip

MCQ

The function $\text{f(x)}=4\sin^3\text{x}-6\sin^2\text{x}+12\sin\text{x}+100$ is strictly:

A
$\text{Increasing in }\Big(\pi,\frac{3\pi}{2}\Big)$
B
$\text{Decreasing in }\Big(\frac{\pi}{2},\pi\Big)$
C
$\text{Decreasing in }\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$
D
$\text{Decreasing in }\Big[0,\frac{\pi}{2}\Big]$

✓

Answer

$\text{Decreasing in }\Big(\frac{\pi}{2},\pi\Big)$

Solution:

We have, $\text{f(x)}=4\sin^3\text{x}-6\sin^2\text{x}+12\sin\text{x}+100$

$\therefore\ \text{f(x)}=12\sin^2\text{x}\cdot\cos\text{x}-12\sin\text{x}\cdot\cos\text{x}+12\cos\text{x}$

$=12\cos\text{x}\big[\sin^2\text{x}-\sin\text{x}+1\big]$

$=12\cos\text{x}\big[\sin^2\text{x}+(1-\sin\text{x}\big]$

Now, $1-\sin\text{x}\geq0\text{ and }\sin^2\text{x}\geq0$

$\therefore\ \sin^2\text{x}+\text{f}-\sin\text{x}>0$

Hence, f'(x) > 0, when $\cos\text{x}>0\text{ i.e., x}\in\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$

So, f(x) is increasing when $\text{x}\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)$

and f'(x) < 0, when $\cos\text{x}<0\text{ i.e., x}\in\Big(\frac{\pi}{2},\frac{3\pi}{2}\Big)$

Hence, f(x) is decreasing when $\text{x}\in\Big(\frac{\pi}{2},\frac{3\pi}{2}\Big)$

Hence, f(x) is decreasing in $\Big(\frac{\pi}{2},\pi\Big)$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from APPLICATION OF DERIVATIVES→APPLICATION OF DERIVATIVES — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Consider the following statements:

Statement I: The area bounded by the curve, $\text{y}=\sin\text{x}$ between $\text{x}=0$ and x = 2p is 2 sq. units.

Statement II: The area bounded by the curve, $\text{y}=2\cos\text{x}$ and the x-axis from $\text{x}=0$ to x = 2p is 8 sq. units.

Statement I is true
Statement II is true
Both statements are true
Both statements are false

If $\alpha=\tan^{-1}\Big(\tan\frac{5\pi}{4}\Big)$ and $\beta=\tan^{-1}\Big(-\tan\frac{2\pi}{3}\Big),$ then:

$4\alpha=3\beta$
$3\alpha=4\beta$
$\alpha-\beta=\frac{7\pi}{12}$
$\text{none of these}$

If $f'\left( x \right) = \sin \,\left( {\log \,x} \right)$ and $y = f\,\left( {\frac{{2x + 3}}{{3 - 2x}}} \right)$, then $\frac{{dy}}{{dx}}$ equals

The area of the plane region bounded by the curves $x + 2{y^2} = 0$ and $x + 3{y^2} = 1$ is equal to

Suppose that $f$ is differentiable for all $x$ and that $f '(x) \le 2$ for all x. If $f (1) = 2$ and $f (4) = 8$ then $f (2)$ has the value equal to

${\sin ^{ - 1}}x + {\sin ^{ - 1}}\frac{1}{x} + {\cos ^{ - 1}}x + {\cos ^{ - 1}}\frac{1}{x} = $

Let $f(x) = \int\limits_0^x {{e^{ - {t^2}}}dt} $ for all $x > 0$, then for all $x > 0$

Smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2 is:

$2(\pi-2)$
$\pi-2$
$\pi-1$
$2(\pi+2)$

If $\int_0^\pi {x\,f({{\cos }^2}x + {{\tan }^4}x)\,dx} $ $ = k\int_0^{\pi /2} {f({{\cos }^2}x + {{\tan }^4}x)\,dx,} $ then the value of $k$ is

The value of $\tan\Big\{\cos^{-1}\frac{1}{5\sqrt2}-\sin^{-1}\frac{4}{\sqrt{17}}\Big\}$ is:

$\frac{\sqrt{29}}{3}$
$\frac{29}{3}$
$\frac{\sqrt3}{29}$
$\frac{3}{29}$